FURTHER DEVELOPMENT OF AN OPEN PROBLEM CONCERNING AN INTEGRAL INEQUALITY
WEN-JUN LIU, GUO-SHENG CHENG, AND CHUN-CHENG LI COLLEGE OFMATHEMATICS ANDPHYSICS
NANJINGUNIVERSITY OFINFORMATIONSCIENCE ANDTECHNOLOGY
NANJING210044, CHINA
lwjboy@126.com gshcheng@sohu.com lichunchengcxy@126.com
Received 04 October, 2007; accepted 18 March, 2008 Communicated by F. Qi
ABSTRACT. In this paper, we generalize an open problem posed by Q. A. Ngô et al. in the paper Notes on an Integral Inequality, J. Inequal. in Pure and Appl. Math., 7(4)(2006), Art. 120 and give an affirmative answer to it without the differentiable restriction onf.
Key words and phrases: Integral inequality, Cauchy inequality.
2000 Mathematics Subject Classification. 26D15.
1. INTRODUCTION
Recently, in the paper [6] Ngô et al. studied some very interesting integral inequalities and proved the following result.
Theorem 1.1. Letf(x)≥0be a continuous function on[0,1]satisfying (1.1)
Z 1
x
f(t)dt ≥ Z 1
x
t dt, ∀x∈[0,1].
Then the inequalities (1.2)
Z 1
0
fα+1(x)dx≥ Z 1
0
xαf(x)dx, and
(1.3)
Z 1
0
fα+1(x)dx≥ Z 1
0
xfα(x)dx,
This work was supported by the Science Research Foundation of Nanjing University of Information Science and Technology and the Natural Science Foundation of Jiangsu Province Education Department under Grant No.07KJD510133.
We would like to express deep gratitude to Q. A. Ngô for his helpful comments.
306-07
hold for every positive real numberα >0.
Next, they proposed the following open problem:
Problem 1.1. Letf(x)be a continuous function on[0,1]satisfying (1.4)
Z 1
x
f(t)dt ≥ Z 1
x
t dt, ∀x∈[0,1].
Under what conditions does the inequality (1.5)
Z 1
0
fα+β(x)dx≥ Z 1
0
xαfβ(x)dx, holds forαandβ?
We note that, as an open problem, the condition (1.4) may result in an unreasonable restriction onf(x). We remove it herein and propose another more general open problem:
Problem 1.2. Under what conditions does the inequality (1.6)
Z b
a
fα+β(x)dx≥ Z b
a
(x−a)αfβ(x)dx, hold fora, b, αandβ?
Shortly after the paper [6] was published, Liu et al. [5] gave an affirmative answer to Problem 1.2 for the casea= 0and obtained the following result:
Theorem 1.2. Letf(x)≥0be a continuous function on[0, b], b≥0satisfying (1.7)
Z b
x
fβ(t)dt≥ Z b
x
tβdt, ∀x∈[0, b].
Then the inequality (1.8)
Z b
0
fα+β(x)dx≥ Z b
0
xαfβ(x)dx,
holds for every positive real numberα >0andβ >0.
Almost at the same time, Bougoffa [1] also gave an answer to Problem 1.2 and established the following result (We correct it here according to the presence of the corrigendum in [2]):
Theorem 1.3. Letf(x)≥0be a function, continuous on[a, b]and differentiable in(a, b). If (1.9)
Z b
x
f(t)dt ≥ Z b
x
(t−a)dt, ∀x∈[a, b]
and
f0(x)≤1, ∀x∈(a, b),
then the inequality (1.6) holds for every positive real numberα >0andβ >0.
Very recently, Boukerrioua and Guezane-Lakoud [3] obtained the following result:
Theorem 1.4. Letf(x)≥0be a continuous function on[0,1]satisfying (1.10)
Z 1
x
f(t)dt ≥ Z 1
x
t dt, ∀x∈[0,1].
Then the inequality (1.11)
Z 1
0
fα+β(x)dx≥ Z 1
0
xαfβ(x)dx,
holds forα >0andβ ≥1.
Comparing the above three results, we note that: the condition (1.7) was required in Theorem 1.2, a differentiability condition was restricted onf in Theorem 1.3 whileβ ≥1was demanded in Theorem 1.4. In this paper, we will give an affirmative answer to Problem 1.2 without the differentiable restriction onf by improving the methods of [5], [6] and [3]. Our main result is Theorem 2.1 which will be proved in Section 2.
2. MAINRESULTS AND PROOFS
Theorem 2.1. Letf(x)≥0be a continuous function on[a, b]satisfying (2.1)
Z b
x
fmin{1,β}(t)dt≥ Z b
x
(t−a)min{1,β}dt, ∀x∈[a, b].
Then the inequality (2.2)
Z b
a
fα+β(x)dx≥ Z b
a
(x−a)αfβ(x)dx,
holds for every positive real numberα >0andβ >0.
To prove Theorem 2.1, we need the following lemmas.
Lemma 2.2 ([6], General Cauchy inequality). Letαandβ be positive real numbers satisfying α+β = 1. Then for all positive real numbersxandy, we always have
(2.3) αx+βy ≥xαyβ.
Lemma 2.3. Under the conditions of Theorem 2.1, we have (2.4)
Z b
a
(x−a)αfβ(x)dx≥ (b−a)α+β+1 α+β+ 1 .
Proof. We divide the proof into two steps according to the different intervals ofβ.
Case of0< β≤1: Integrating by parts, we have Z b
a
(x−a)α−1 Z b
x
fβ(t)dt
dx
= 1 α
Z b
a
Z b
x
fβ(t)dt
d(x−a)α
= 1 α
(x−a)α Z b
x
fβ(t)dt x=b
x=a
+ 1 α
Z b
a
(x−a)αfβ(x)dx
= 1 α
Z b
a
(x−a)αfβ(x)dx.
which yields (2.5)
Z b
a
(x−a)αfβ(x)dx=α Z b
a
(x−a)α−1 Z b
x
fβ(t)dt
dx.
On the other hand, by (2.1), we get Z b
a
(x−a)α−1 Z b
x
fβ(t)dt
dx
≥ Z b
a
(x−a)α−1 Z b
x
(t−a)βdt
dx
= 1
β+ 1 Z b
a
(x−a)α−1
(b−a)β+1−(x−a)β+1 dx
= (b−a)α+β+1 α(α+β+ 1). Therefore, (2.4) holds.
Case ofβ >1: We note that the following result has been proved in the first case (2.6)
Z b
a
(x−a)αf(x)dx≥ (b−a)α+2 α+ 2 . Using Lemma 2.2, we get
(2.7) 1
βfβ(x) + β−1
β (x−a)β ≥f(x)(x−a)β−1.
Multiplying both sides of (2.7) by(x−a)α and integrating the resultant inequality fromatob, we obtain
(2.8)
Z b
a
(x−a)αfβ(x)dx+ (β−1) Z b
a
(x−a)α+βdx≥β Z b
a
(x−a)α+β−1f(x)dx,
which implies (2.9)
Z b
a
(x−a)αfβ(x)dx+ β−1
α+β+ 1(b−a)α+β+1 ≥β Z b
a
(x−a)α+β−1f(x)dx.
Moreover, by using (2.6), we get (2.10)
Z b
a
(x−a)αfβ(x)dx+ β−1
α+β+ 1(b−a)α+β+1 ≥ β
α+β+ 1(b−a)α+β+1,
which implies (2.4).
We now give the proof of Theorem 2.1.
Proof of Theorem 2.1. Using Lemma 2.2 again, we obtain
(2.11) β
α+βfα+β(x) + α
α+β(x−a)α+βdx≥(x−a)αfβ(x)dx, which gives
(2.12) β
Z b
a
fα+β(x)dx+α Z b
a
(x−a)α+βdx≥(α+β) Z b
a
(x−a)αfβ(x)dx.
Moreover, by using Lemma 2.3, we get (α+β)
Z b
a
(x−a)αfβ(x)dx=α Z b
a
(x−a)αfβ(x)dx+β Z b
a
(x−a)αfβ(x)dx
≥α(b−a)α+β+1 α+β+ 1 +β
Z b
a
(x−a)αfβ(x)dx,
that is (2.13) β
Z b
a
fα+β(x)dx+α(b−a)α+β+1
α+β+ 1 ≥α(b−a)α+β+1 α+β+ 1 +β
Z b
a
(x−a)αfβ(x)dx,
which completes the proof.
REFERENCES
[1] L. BOUGOFFA, Note on an open problem, J. Inequal. Pure Appl. Math., 8(2) (2007), Art. 58.
[ONLINE:http://jipam.vu.edu.au/article.php?sid=871].
[2] L. BOUGOFFA, Corrigendum of the paper entitled: Note on an open problem, J. Inequal. Pure Appl. Math., 8(4) (2007), Art. 121. [ONLINE:http://jipam.vu.edu.au/article.php?
sid=910].
[3] K. BOUKERRIOUA AND A. GUEZANE-LAKOUD, On an open question regarding an integral inequality, J. Inequal. Pure Appl. Math., 8(3) (2007), Art. 77. [ONLINE:http://jipam.vu.
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